# Derivative of a matrix trace w.r.t complex matrix

I have got the derivative of a trace w.r.t a real matrix as follows $\frac {\partial}{\partial \mathbf X} tr[\mathbf{(X^TCX)}^{-1}(\mathbf{X^T BX})] = -2 \mathbf{CX(X^TCX)^{-1}X^TBX(X^TCX)^{-1}} + 2\mathbf{BX(X^TCX)^{-1}}$

where $\mathbf {B,C}$ is symmetric.

Now I want to solve the trace w.r.t a complex matrix as follows

$\frac {\partial}{\partial \mathbf X} tr[\mathbf{(X^H CX)}^{-1}(\mathbf{X^H BX})]$

where $\mathbf{B,C}$ conjugate symmertic.

I will feel very gratful if anyone leave me some tips.

• I forgot to say that $\mathbf X$ is not a square matrix. – freddy Nov 10 '16 at 13:24
• Is it right that you interpret $A:=\partial/(\partial X) tr[\ldots]$ as linear form on the space $\mathbb{R}^{n\times m}\ni X$ (i.e., $A$ applied to a matrix $\delta X\in \mathbb{R}^{n\times m}$ is $\sum_{(i,j)\in \{1,\ldots,n\}\times\{1,\ldots,m\}} A_{i,j} \cdot \delta X_{i,j}$)? – Tobias Nov 10 '16 at 13:42

Maybe not the final solution but a step forward:

Following calculation is valid only under the assumption that $X$ is square and regular.

You can simplify $(X^H CX)^{-1}X^H BX = X^{-1}C^{-1}X^{-H} X^H B X = X^{-1}C^{-1}B X$.

The Gateaux-derivative $\delta A:=\left.\partial_\varepsilon A(X+\varepsilon \delta X)\right|_{\varepsilon=0}$ of the inverse-matrix operation $A(X):=X^{-1}$ results from

\begin{align*} 1 &= A(X) X,\\ 0 &= \delta(AX) = \delta A\, X + A\,\delta X,\\ \delta A &= -A\,\delta X\,X^{-1}\\ &= \underline{\underline{X^{-1}\,\delta X\,X^{-1}}}. \end{align*}

This helps with the calculation of the Gateaux-derivative of your original equation: \begin{align*} \delta \operatorname{tr}[X^{-1}C^{-1}B X] &= \operatorname{tr}[X^{-1} \delta X X^{-1} C^{-1}B X + X^{-1}C^{-1}B \delta X] \end{align*}

• I forgot to say that $\mathbf X$ is not a square matrix. – freddy Nov 10 '16 at 13:24
• @freddy The general case where $X$ is not square can be treated analogously. It is just more work. $\delta( X^H ) = (\delta X)^H$. – Tobias Nov 10 '16 at 14:02